TCS NQT Placement Papers and Questions

Let X be the number of 1 rupee and Y be the number of 2 rupees coin

X+Y= 100

Take x = 100, Y=0

Likewise,Y=1, X=98

It will go till Y=50, X=0

Therefore there are 51 ways

The car moves 1/2 mile in 1 minute.

Speed of the car = 1/2 mile for 1 min.

30 miles for 60 min(1 hr).

Therefore, the speed of the car is 30 mph.

Also, the speed of the observant car is 55 mph.

And both are moving in the same direction, we must add the speeds.

So, speed of the car = (55 + 30) mph

= 85 mph.

We are given to select the next number in the following series :

70, 54, 45, 41, . . .

Let a(n) denote the n-th term of the given series.

Then, we observe that

$a(1)=70,\\\\a(2)=54=70-16=a(1)-4^{2},\\\\a(3)=45=54-9=a(2)-3^{2},\\\\a(4)=41=45-4=a(3)-2^{2},$

Repeating the above rule, we get

$a(5)=a(4)-1^{2}=41-1=40.$

Thus, the next term in the given series is 40.

George can do the work in 8 hours

⇒ 1 hour = 1/8 of the work

Paul can do the work in 10 hours

⇒1 hour = 1/10 of the work

Hari can do the work in 12 hours

⇒ 1 hour = 1/12 of the work

Find the amount of work that can be completed in an hour:

⇒ 1 hour = 1/8 + 1/10 + 1/12 = 37/120

Find the amount of work completed in 2 hours (9 am to 11 am):

⇒ 1 hour = 37/120

⇒2 hours = 37/120 x 2 = 37/60

Find the amount of work remain:

Work remain = 1 - 37/60 = 23/60

Find the amount of work Paul and Hari can do in an hour:

1 hour = 1/10 + 1/12 = 11/60

Find the number of hours needed for them to complete the rest of the work:

Number of hours needed = 23/60 ÷ 11/60 = 2.09 hours = 2 hours

Find the time the work is completed:

11am + 2 hours = 1 pm

Shush completes work in s days.

so in one day $\frac{1}{s}$th part of the day.

Similarly;

Komal can complete $\frac{1}{k}$th part of the day.

$\frac{1}{s} + \frac{1}{k} = \frac{1}{7}$

Shush is $\frac{7}{4}$times more efficient

therefore, $\frac{1}{s} = \frac{7}{4}\times \frac{1}{k}$

$$\frac{4}{7}\times \frac{1}{s} = \frac{1}{k}[\latex]

so, last equation becomes

$$\frac{1}{s} + \frac{4}{7}\times = \frac{1}{7}[\latex]

S = 11 Days

Acc to the question, there are 3 equations given:

a+b+ab= 8-----(i)

b+c+bc=15------(ii)

c+a+ca = 35---------(iii)

Now adding 1 to both the sides of equ (i) we get,

a+b+ab+1= 8+1

or, (a+1)(b+1)=9.

Similarly we get, (b+1)(c+1)=16 and (a+1)(c+1)=36.

now starting from the second equation,

We know that 16 has many factors but we have to brute force for satisfying the equation.

We know, 16 = 2 * 8 = (1+1) * (7+1) now, comparing it with equ(ii) we get, b=1,c=7.

Now putting the value of c in any two of the remaining equation we get,

(a+1)*(c+1)=36

or, (a+1)*(7+1)=36

or,(a+1) = 36/8 = 4.5

or, a = 4.5 - 1 = 3.5.

Now, a+b+c+abc = 3.5 + 1 + 7 + 3.5*1*7 = 36.

Given,

First-term = a

third term = b

sixth term = c

We see that Each alphabet repeats for the number of times its position in the English Alphabet,

This series will come to end at z repeating itself for 26 times.

Last term z will be $$\frac{26\times 27}{2}[\latex] = 350th term

We can say that ,

N th alphabet will be existing from Σ(n-1)+1 to ΣN

ΣN ≥ 260

Let n = 23

Σ(n-1) +1 = 254

Σn = 276

Therefore , 23rd Alphabet will exist from 254 to 276 terms .

23rd Alphabet is W

The 260th term of the series is W

Here,
Two things have to be taken into account:
• Common difference
• No. of such pairs

For simplicity,
Consider range as 1 - 10
Here, when the common difference exceeds 5, numbers are not paired as per the condition.

Like if you take 6 as common difference, not all the pairs will have same value, because it is impossible.

•°• For range 1 - n, we have (n/2) common differences.

Now,
Coming to the given range 1 - 70
n = 70
=> (n/2) = 35

•°• No. of common differences that will satisfy the condition = 35

Now, Find sum of these differences.
1 + 2 + 3 + …… + 35
= (35 × 36) ÷ 2
= 630

•°• Required sum of all such common differences = 630

Let r be half the side of the square.

Consider one of the twelve triangle segments made by a side of the dodecagon and the centre of the dodecagon ( = the centre of the square).

The angle at the centre is 360° / 12 = 30°.

The two sides of the triangle that meet at the centre each have length r, as this is the radius of the circumscribed circle.

The area of this small triangle is then (1/2) r² sin 30° = r² / 4.

The dodecagon is made up of twelve of these triangles, so

area(dodecagon) = 12 × r² / 4 = 3 r².

As r is half the side of the square,

area(square) = 4 r².

Therefore

area(dodecagon ) / area(square) = 3 / 4

=> area(dodecagon) = 3 × 24 / 4 = 3 × 6 = 18

Here we have to find out the number of ways we can distribute 100 rupees money to rupee 1 and rupee 2.

The simplest method is to find out the number of ways is to calculate the number of coins we use to break partitions. As an example, the ways are

1, 1, 1,..........................................,1 (100 times 1)

1, 1, 1, ......................................, 1 and 2 (98 times 1 and 1 time 2)

1, 1, 1, ....................................., 1, 2 , 2 ( 96 times 1 and 2 times 2)

.............................

proceeding in this way, our last combination for taking the coins be

2, 2, 2, ........................, 2 (50 times 2)

So, here, we can find the numbers of ways by counting the number of 2 we took and is equal to 50 + 1 = 51.

Answer: 51

By backtracking

Option a)

The password would be 8653513649

8653513649 ÷ 11 = 786683059

8653513649 ÷ 8 = 1081689206 Remainder 1

⇒ Number cannot be divisible by 8

Option b)

The password would be 7653513648

7653513648 ÷ 11 = 695773968

7653513648 ÷ 8 = 956689206

⇒ The number is both divisible by 8 and 11

Option c)

The password would be 3653513644

3653513644 ÷ 11 = 332137604

3653513644 ÷ 8 = 456689205 Remainder 4

⇒ The number cannot be divisible by 8

Option d)

The password would be 2653513647

2653513647 ÷ 11 = 241228513 Remainder 4

2653513647 ÷ 8 = 331689205 Remainder 7

⇒ The number cannot be divisible by 8 and 11

Answer: (b) 7, 8

Find the circumference of the front wheel:

Circumference = 2πr

Circumference = 2π(28) = 56π unsyms

Find the total distance covered by the front wheel:

1 revolution = 56π unsyms

54 revolutions = 56π x 54 = 3024π unsyms

Find the circumference of the back wheel:

Circumference = 2πr

Circumference = 2π(14) = 28π unsyms

Find the number of revolutions made by the back wheel:

Number of revolutions = 3024π ÷ 28π = 108

Step-by-step explanation:

first of all a+b+c=3000 --------(1)eqn

second: 2/3(b+c)=a; -----------(2)eqn

third: (a+b)/2=c; -----------------(3)eqn

so we got 3 equation and 3 variable

a+b+c=3000; (1)eqn

a+b-2c=0; (3)eqn

subtracting both we get c=1000

The probability that Anu will get to watch the Wimbledon match is 3/8

Step-by-step explanation:

Probability of any number appearing = 1/4

Anu wins in the following condition

1) If 1 occus on the first roll then there is no chance of her winning because in the next roll the number appeaing will be either 1 or greater than 1

2) If 2 appears on 1st roll then Anu will win if 1 appears on the second roll

3) If 3 appears on the 1st roll then Anu will win if 1 or 2 appears on the second roll

4) If 4 appears on the first roll then Anu will win if 1, 2, 3 appears on the second roll

Therefore, the probability of Anu winning

$$\frac{1}{4}\times \frac{1}{4} + \frac{1}{4}\times \frac{2}{4} + \frac{1}{4}\times \frac{3}{4} = \frac{3}{8}[\latex]